📚 A-Level Maths: Finding Stationary Points and Determining Extrema | A-Level数学：驻点的求法与极值判定
Stationary points are one of the most frequently tested topics in A-Level mathematics, appearing in both pure mathematics and applied contexts. Understanding how to find them and correctly classify them as local maxima, local minima, or points of inflection is essential for exam success.
驻点是A-Level数学中最高频的考点之一，既出现在纯数学部分，也出现在应用问题中。掌握驻点的求法并正确判定其为极大值、极小值还是拐点，是考试取得高分的关键。
1. What Is a Stationary Point? | 什么是驻点？
A stationary point of a function f(x) is a point where the derivative f'(x) equals zero. At such a point, the gradient of the tangent to the curve is horizontal, meaning the function is momentarily not increasing or decreasing.
函数 f(x) 的驻点是指导数 f'(x) 等于零的点。在驻点处，曲线切线的斜率为零，函数在该点瞬间既不增加也不减少。
To find stationary points, we solve the equation:
求驻点需要解方程：
f'(x) = 0
For example, for f(x) = x³ - 3x + 1, we have f' - 3. Setting 3x² - 3 = 0 gives x = ±1, so the stationary points occur at x = 1 and x = -1.
例如，对于 f(x) = x³ - 3x + 1，f'(x) = 3x² - 3。令 3x² - 3 = 0，得 x = ±1，因此驻点出现在 x = 1 和 x = -1 处。
2. The First Derivative Test | 一阶导数判定法
Once the stationary points are found, we need to determine their nature. The first derivative test examines the sign of f'(x) immediately to the left and right of the stationary point.
找到驻点后，需要判断其性质。一阶导数判定法通过观察驻点左右两侧 f'(x) 的符号变化来判断。
If f'(x) changes from positive to negative, the point is a local maximum.
如果 f'(x) 从正变负，则该点为局部极大值。
If f'(x) changes from negative to positive, the point is a local minimum.
如果 f'(x) 从负变正，则该点为局部极小值。
If f'(x) does not change sign, the point is a stationary point of inflection.
如果 f'(x) 符号不变，则该点为驻型拐点。
Let us apply this to f(x) = x³ - 3x + 1. Around x = -1, f'(x) = 3x² - 3. For x = -1.1, f'(-1.1) = 3(1.21) - 3 = 0.63 > 0; for x = -0.9, f' 57 < 0. Since the derivative changes from positive to negative, x = -1 is a local maximum.
以 f(x) = x³ - 3x + 1 为例。在 x = -1 附近，f'(x) = 3x² - 3。取 x = -1.1，f'(-1.1) = 3(1.21) - 3 = 0.63 > 0；取 x = -0.9，f'(-0.9) = 3(0.81) - 3 = -0.57 < 0。导数从正变负，因此 x = -1 是局部极大值。
Similarly, around x = 1: f'(0.9) = 3(0.81) - 3 = -0.57 < 0, and f' = 0.63 > 0. The derivative changes from negative to positive, so x = 1 is a local minimum.
同理，在 x = 1 附近：f'(0.9) = 3(0.81) - 3 = -0.57 < 0，f'(1.1) = 3(1.21) - 3 = 0.63 > 0。导数从负变正，因此 x = 1 是局部极小值。
3. The Second Derivative Test | 二阶导数判定法
The second derivative test is often faster. At a stationary point where f'(x) = 0:
二阶导数判定法通常更快。在 f'(x) = 0 的驻点处：
If f''(x) > 0, the point is a local minimum.
若 f''(x) > 0，该点为局部极小值。
If f''(x) < 0, the point is a local maximum.
若 f''(x) < 0，该点为局部极大值。
If f''(x) = 0, the test is inconclusive, and the first derivative test must be used.
若 f''(x) = 0，该判定法失效，必须使用一阶导数判定法。
For f(x) = x³ - 3x + 1, f''(x) = 6x. At x = -1, f''(-1) = -6 < 0, confirming a local maximum. At x = 1, f''(1) = 6 > 0, confirming a local minimum.
对于 f(x) = x³ - 3x + 1，f''(x) = 6x。在 x = -1 处，f''(-1) = -6 < 0，确认是局部极大值；在 x = 1 处，f''(1) = 6 > 0，确认是局部极小值。
4. Classifying Stationary Points with a Table | 用表格判定驻点类型
In exams, you are often expected to present your classification clearly. A sign table or table of values is the most systematic way.
考试中通常要求清晰地展示判定过程。符号表或数值表是最系统的方法。
For example, consider f(x) = 2x³ - 9x² + 12x - 3. Then f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2). The stationary points are x = 1 and x = 2.
例如，考虑 f(x) = 2x³ - 9x² + 12x - 3。则 f'(x) = 6x² - 18x + 12 = 6(x - 1)(x - 2)。驻点为 x = 1 和 x = 2。
| x | x < 1 | x = 1 | 1 < x < 2 | x = 2 | x > 2 |
| f'(x) | positive | 0 | negative | 0 | positive |
| Slope | rising | horizontal | falling | horizontal | rising |
| Conclusion | — | maximum | — | minimum | — |
The table clearly shows that x = 1 gives a local maximum and x = 2 gives a local minimum.
表格清晰表明 x = 1 为局部极大值，x = 2 为局部极小值。
5. Points of Inflection | 拐点
A point of inflection is a point where the curve changes its concavity, i.e. where the second derivative changes sign. If the curve also has a horizontal tangent at that point, it is called a stationary point of inflection.
拐点是曲线凹凸性发生改变的点，即二阶导数符号改变的点。如果该点处的切线也是水平的，则称为驻型拐点。
For example, f(x) = x³ has f'(x) = 3x² and f''(x) = 6x. At x = 0, f'(0) = 0 and f''(0) = 0. The first derivative test: for x < 0, f'(x) > 0; for x > 0, f'(x) > 0. Since the sign does not change, (0, 0) is a stationary point of inflection.
例如，f(x) = x³ 的 f'(x) = 3x²，f''(x) = 6x。在 x = 0 处，f'(0) = 0 且 f''(0) = 0。一阶导数判定法：x < 0 时 f'(x) > 0；x > 0 时 f'(x) > 0。符号未改变，因此 (0, 0) 是驻型拐点。
Note that a non-stationary point of inflection occurs when f''(x) = 0 but f' at x = 0 for f(x) = x³ + x. The curve changes concavity but the tangent is not horizontal.
注意，非驻型拐点出现在 f''(x) = 0 但 f'(x) ≠ 0 的情况，例如 f(x) = x³ + x 在 x = 0 处。曲线凹凸性改变，但切线不是水平的。
6. Worked Example: Complete Classification | 综合例题：完整分类
Find and classify all stationary points of the function f(x) = x⁴ - 4x³ + 4x².
求函数 f(x) = x⁴
Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图